| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Topic | Completing the square and sketching |
| Type | Tangent to curve: find parameter |
| Difficulty | Standard +0.3 This is a straightforward multi-part question combining completing the square (routine algebraic manipulation), identifying symmetry (direct read-off), and finding a tangent condition (standard discriminant method). All techniques are well-practiced at A-level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points1.07m Tangents and normals: gradient and equations |
**Question 2(i), 2(ii), 2(iii)**
**(i)**
- $2(x+\frac{3}{2})^2 + \frac{1}{2}$ **B3** **B1** for each of $p=2$, $q=\frac{3}{2}$, $r=\frac{1}{2}$
**(ii)**
- $x = -\frac{3}{2}$ **B1** FT: State $x = -\frac{3}{2}$ (ft on their $q$)
**(iii)**
- $2x^2 + 6x + 5 = k - 2x$, rearranging to $2x^2 + 8x + 5 - k = 0$ **B1** Equate and rearrange to obtain $2x^2 + 8x + 5 - k = 0$
- $64 - 4 \times 2 \times (5-k) = 0$, giving $24 + 8k = 0$ **M1** Use $b^2 - 4ac = 0$ to attempt to find $k$
- $k = -3$ **A1**
OR
- $4x + 6 = -2$ **B1** Differentiate and equate to get $4x+6=-2$
- $x=-2$, $y=1$; $1 = k+4$ **M1** Solve their $\frac{dy}{dx}=-2$ to find point of intersection and hence attempt to find $k$
- $k = -3$ **A1**2 (i) Express $2 x ^ { 2 } + 6 x + 5$ in the form $p ( x + q ) ^ { 2 } + r$.\\
(ii) State the equation of the line of symmetry of the curve $y = 2 x ^ { 2 } + 6 x + 5$.\\
(iii) Find the value of the constant $k$ for which the line $y = k - 2 x$ is a tangent to the curve $y = 2 x ^ { 2 } + 6 x + 5$.
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2018 Q2 [11]}}